Question

For two complex numbers $z_1$ and $z_2;(a{z}_1+b\overline{z_1})(c{z}_2+d\overline{z_2})=(c{z}_1+d\overline{z_1})(a{z}_2+b\overline{z_2})$ $b\ne 0,d\ne 0$ if

• A. $\frac{a}{b}=\frac{c}{d}$
• B. $\frac{a}{d}=\frac{b}{c}$
• C. $|z_1|=|z_2|$
• D. $\arg \left(z_1\right)=\arg \left(z_2\right)$

Answer 1

Answer: (A), (D)

The correct options are
A $\frac{a}{b}=\frac{c}{d}$
D $\arg \left(z_1\right)=\arg \left(z_2\right)$

$ac{z}_1{z}_2+ad{z}_1\overline{z_2}+bd\overline{z_1}\overline{z_2}+bc\overline{z_1}{z}_2$

$=ac{z}_1{z}_2+ad\overline{z_1}{z}_2+bc{z}_1\overline{z_2}+bd\overline{z_1}\overline{z_2}$

$\Rightarrow (ad-bc)(z_1\overline{z_2}-\overline{z_1}{z}_2)=0$

$\Rightarrow ad-bc=0\Rightarrow \frac{a}{b}=\frac{c}{d}$

if $z_1\overline{z_2}-\overline{z_1}{z}_2=0$

$\Rightarrow \frac{|z_1|e^{i{\theta }_1}}{|z_2|e^{-i{\theta }_1}}=\frac{|z_2|e^{i{\theta }_2}}{|z_2|e^{-i{\theta }_2}}\Rightarrow {\theta }_1={\theta }_2$